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Block 1 , with mass \(m_{1}\) and speed \(3.0 \mathrm{~m} / \mathrm{s}\), slides along an \(x\) axis on a frictionless floor and then undergoes a one-dimensional elastic collision with stationary block 2 , with mass \(m_{2}=0.40 m_{1}\). The two blocks then slide into a region where the coefficient of kinetic friction is \(0.50\); there they stop. How far into that region do (a) block 1 and (b) block 2 slide?

Short Answer

Expert verified
Block 1 slides 0.0188 m and block 2 slides 0.4688 m.

Step by step solution

01

Apply conservation of momentum

In an elastic collision, both momentum and kinetic energy are conserved. Let the initial velocity of block 1 be \(u_1 = 3.0\, \mathrm{m/s}\) and the initial velocity of block 2 be \(u_2 = 0\, \mathrm{m/s}\). The equation for conservation of momentum is:\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]Substituting \(m_2 = 0.40 m_1\) and \(u_2 = 0\),\[ m_1 \times 3.0 = m_1 v_1 + 0.4 m_1 v_2 \] Simplifies to: \[ 3.0 = v_1 + 0.4 v_2 \]
02

Apply conservation of kinetic energy

For an elastic collision, kinetic energy is conserved. The equation for conservation of kinetic energy is:\[\frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2\]Substituting \(u_2 = 0\) and \(m_2 = 0.40 m_1\),\[ \frac{1}{2} m_1 (3.0)^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} \times 0.40 m_1 v_2^2 \]Simplifying this equation gives:\[ 9.0 = v_1^2 + 0.40 v_2^2 \]
03

Solve simultaneous equations

We have two equations:1. \(3.0 = v_1 + 0.4 v_2\)2. \(9.0 = v_1^2 + 0.40 v_2^2\)We solve these equations simultaneously to find \(v_1\) and \(v_2\). By substituting \(v_1 = 3.0 - 0.4 v_2\) into the second equation and solving, we find \(v_1 = 0.4286\, \mathrm{m/s}\) and \(v_2 = 2.1429\, \mathrm{m/s}\).
04

Determine distance using kinematics

After the collision, both blocks slide into a region with kinetic friction \(\mu_k = 0.50\). The friction decelerates the blocks until they stop. Using the work-energy principle:\[ \text{friction work} = \text{initial kinetic energy} \]For block 1: \(f_k = \mu_k m_1 g\), and using \(v_1 = 0.4286\, \mathrm{m/s}\),\[ \mu_k m_1 g d_1 = \frac{1}{2} m_1 v_1^2 \]\[ d_1 = \frac{v_1^2}{2\mu_k g} = \frac{(0.4286)^2}{2\times 0.50 \times 9.8} \approx 0.0188 \mathrm{~m} \]For block 2:\[ d_2 = \frac{v_2^2}{2\mu_k g} = \frac{(2.1429)^2}{2\times 0.50 \times 9.8} \approx 0.4688 \mathrm{~m} \]
05

Conclusion

The calculations indicate that (a) block 1 slides approximately 0.0188 meters and (b) block 2 slides approximately 0.4688 meters into the region with kinetic friction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum
In physics, the principle of conservation of momentum is fundamental, especially when dealing with collisions. The conservation of momentum states that the total momentum of a closed system is constant when no external forces act upon it. This concept is particularly useful in analyzing elastic collisions, where kinetic energy is also conserved.

In the exercise, we observe an elastic collision between two blocks. The initial momentum is the sum of the product of each block's mass and its velocity. For block 1 with mass \( m_1 \) moving at 3.0 m/s, and block 2 stationary, the initial equation is \( m_1 \times 3.0 + 0 \). After the collision, both blocks are moving with new velocities \( v_1 \) and \( v_2 \), maintaining the equation:

\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Replacing \( m_2 \) with \( 0.40 m_1 \) simplifies the equation further. This derivation emphasizes how momentum is redistributed between the blocks, a key feature of collisions.
Conservation of Kinetic Energy
In an elastic collision, not only is momentum conserved, but kinetic energy as well. This means the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The conservation of kinetic energy is expressed by the equation:

\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
With block 2 initially stationary, and substituting \( m_2 = 0.40 m_1 \), we arrive at:

\[ \frac{1}{2} m_1 (3.0)^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} \times 0.40 m_1 v_2^2 \]
Here's how it simplifies and helps us understand the relationship between the velocities of the two blocks. This principle allows us to solve for the unknown velocities after the collision. Understanding this concept is crucial for predicting how energy transforms and transports within a system.
Kinetic Friction
Kinetic friction comes into play once the blocks begin to move across a surface after the collision. It is a force that opposes motion, acting parallel to the contact surface. This frictional force depends on the normal force and a coefficient of kinetic friction, \( \mu_k \).

When the blocks enter this region with \( \mu_k = 0.50 \), kinetic friction performs work on the blocks, gradually slowing them. The frictional force \( f_k \) is calculated as:

\[ f_k = \mu_k \times m \times g \]
In this scenario, both blocks experience deceleration due to kinetic friction, and eventually stop. The distance each block travels in this frictional zone can be calculated using their initial kinetic energies.
Work-Energy Principle
The work-energy principle connects the concepts of work and kinetic energy, providing essential insights into how forces lead to motion. It states that the work done by all forces acting on an object equals the change in its kinetic energy. This principle is extremely useful in calculating how far objects will move under the influence of forces like friction.

After the collision, the blocks have certain kinetic energies. As they slide over a surface with kinetic friction, the work done by this friction equals the initial kinetic energy they had. For each block, the work done by friction is given by:

\[ \text{friction work} = \text{initial kinetic energy} \]
For block 1, this becomes \( \mu_k m_1 g d_1 = \frac{1}{2} m_1 v_1^2 \). Solving for \( d_1 \) provides the distance block 1 travels before stopping. Similarly, the distance for block 2 is found using \( v_2 \). This framework is helpful to understand how objects lose energy over time due to friction.

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